Q.1. The pH of a 0.1 molar solution of the acid. HQ is 3. The value of the
ionization constant, Ka of this acid is : (1) 1 × 10–5. (2) 1 × 10–7. (3) 3 × 10–1.
Q.1
Ans. Sol.
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is : (1) 1 × 10–5 (2) 1 × 10–7 (3) 3 × 10–1 (4) 1 × 10–3 [1] pH = 3 [H+] = 10 – 3 M = C
Sol.
a P 2 ( V b) RT V Reduce to P(V –b) = RT or PV = RT + bP
10 3 10 3 = = = 0.01 E°X Hence, Y = Ni and X = Zn
Ans.
CH3–CH–OH iso propyl alcohol it will give CH3 iodoform
TS H RT
Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be : (1) 240 pm (2) 152 pm (3) 75 pm (4) 300 pm [2]
Ethyl methyl ketone O CH 3–C–CH 2– CH3 It will give
[1] G° = H° – TS° = – RT ln K ln K =
H
C=C H CH3–CH2–CH3 Trans -2-hexene Q.14
Ans. Sol.
Sol.
3 a = 4r
3 351 = 152 pm Ans. 4
How many chiral compounds are possible on monochlorination of 2-methyl butane ? (1) 4 (2) 6 (3) 8 (4) 2 [1]
CH3–CH–CH2–CH3 CH3 Cl2/hv
H * CH3–CH–CH–CH3 + CH3–C–CH2–CH3 * CH2–Cl CH3 2 form 2 form (Enantiomer) (Enantiomer) Cl
Total 4 Chiral molecules
Q.19
Ans. Sol.
Kf for water is 1.86 K kg mol–1. If you automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8°C ? (1) 39 g (2) 27 g (3) 72 g (4) 93 g [4] Tf = Kf × m or 2.8 = 1.86 ×
W / 62 1
Q.22
Ans.
Ortho-Nitrophenol is less soluble in water than p- and m-Nitrophenols because : (1) o-Nitrophenol shows intermolecular Hbonding (2) Melting point of o-Nitrophenol is lower than those of m-and p-isomers. (3) o-Nitrophenol is more volatile in steam than those of m-and p-isomers. (4) o-Nitrophenol shows intramolecular Hbonding [4] OH
W = 93
NO2
Q.20
Ans.
Sol.
In which of the followng pairs the two species are not isostructural ? (1) PF5 and BrF5 (2) AlF63– and SF6 (3) CO32– and NO3– (4) PCl4+ and SiCl4 [1]
F F P F F F PF5 3 sp d–T.B.P
F
F F
Sol.
Due to intramolecular H-bonding its H-bonding with H2O decreases. Solubility became less.
Q.23
In the given transformation, which of the following is the most appropriate reagent ?
Br
F
CH = CHCOCH3
F
Re agent
HO CH = CHCH 2CH 3
Square pyramidal 3 2 sp d
HO
(1) Na, Liq. NH3 Q.21
Ans. Sol.
For a first order reaction, (A) products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M, is (1) 3.47 × 10–5 M/min (2) 1.73 × 10–4 M/min (3) 1.73 × 10–5 M/min (4) 3.47 × 10–4 M/min [4] 1/ 2 1/ 2 0.1 M t 0.05 M t 0.025 M t1/2 = 40/2 = 20 min
0.693 Now, r = K [A] = 20 min × 0.01
= 3.47 × 10–4 M/ min Ans.
(2) NaBH4
–
Ans. Sol.
(3) NH2NH2, OH [3]
(4) Zn – Hg/HCl
O CH=CH–C–CH3 Reagnet HO
CH=CH–CH2–CH3
HO
Reagent should not effect – OH and C=C Alkaline medium is best suited Wolf - Khischner NH2NH2/ OH is most appropriate.
Q.24
According to Freundlich adsorption isotherm, which fo the following is correct ? (1)
x p1 / n m
Q.27
Ans.
x p0 (2) m
Ans.
x p1 m (4) All the above the correct for different ranges of pressure. [1]
Sol.
According to Freundlich ,
Q.25
The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is : (1) 1.02 M (2) 2.05 M (3) 0.50 M (4) 1.78 M [2] Molarity
(3)
Ans. Sol.
x P1/ n m Note : value of n is constant for a particular system from Freundlich isotherm.
n solute 120 / 60 1000 = V (1000 120) / 1.15 solution 2.05 M Ans.
Q.26
Ans. Sol.
Which of the following on thermal decomposition yields a basic as well as an acidic oxide ? (1) CaCO3 (2) NH4NO3 (3) NaNO3 (4) KClO3 [1] CaCO3 CaO + CO2 Basic Acidic
Aspirin is known as : (1) Acetyl salicylate (2) Methyl salicylic acid (3) Acetyl salicylic acid (4) Phenyl salicylate [3] COOH O O–C–CH3
Sol. Aspirin Acetyl Salicylic Acid
Q.28
Ans. Sol.
Q.29
Ans. Sol. Q.30
Ans. Sol.
Which of the following compounds can be detected by Molisch's test ? (1) Amines (2) Primary alcohols (3) Nitro compounds (4) Sugars [4] Molisch Test is for Carbohydrates Sugar can be detected by Molisch Test. What is DDT among the following : (1) Biodegradable pollutant (2) Non-biodegradable pollutant (3) Greenhouse gas (4) A fertilizer [2] DDT is Non-biodegradable pollutant Very pure hydrogen (99.9%) can be made by which of the following processes ? (1) Electrolysis of water (2) Reaction of salt like hydrides with water (3) Reaction of methane with steam (4) Mixing natural hydrocarbons of high molecular weight [2] M H + H 2O MOH + H2 (Pure)
Q.31
Let aˆ and bˆ be two unit vectors. If the vectors c aˆ 2bˆ and d 5aˆ 4bˆ are perpendicular to each other, then the angle between aˆ and bˆ (1*)
6
(3) Ans. Sol.
3
(2)
4
(4)
2
[1] c·d = 0 (a 2 b) · (5a 4b ) = 0 5 | a | 2 6a b 8 | b | 2 = 0 5 + 6 · 1 · 1 cos – 8 = 0 cos =
Q.33
Ans. Sol.
Consider the function f(x) = | x – 2 | + | x – 5 |, x R. Statement 1 : f '(4) = 0 Statement 2 : f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5). (1*) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 2. (2) Statement 1 is true, Statement 2 is false. (3) Statement 1 is false, Statement 2 is true. (4) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. [1] f(x) = | x – 2 | + | x – 5 | , x R 2 x 7, f(x) = 3, 2 x 7,
3 1 = = . 6 2 3
x2 2x5 x5 y
7
Q.32
If the integral 5 tan x
tan x 2 dx
Ans. Sol.
3
= x + a ln | sin x – 2cos x | + k
then a is equal to (1) 1 (2*) 2 (3) –1 (4) – 2 [2] Differentiating both sides,
O
a=
5
It is clear that f(x) is continuous in R and differentiable in (– , 2) (2, 5) (5, ) Statement 2 is correct. Statement 1 is also correct but Statement 2 is not the correct explanation of Statement 1.
5 tan x a = 1+ (cos x + 2sinx) tan x 2 sin x 2 cos x 5 sin x a (cos x 2 sin x ) =1 sin x 2 cos x sin x 2 cos x 5sin x – a (cos x + 2sin x) = sin x – 2cos x 4sin x + 2cos x = a (cos x + 2sin x)
x 2
Q.34
If the line 2x + y = k passes through the point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2, then k equals (1*) 6
2 ( 2 sin x cos x ) a = 2. cos x 2 sin x
(3) Ans. Sol.
29 5
(2)
11 5
(4) 5
[1] Since, M divides A & B in the ratio 3 : 2. Coordinates of M are 6 2 12 2 8 , 14 , 5 5 5 5
3 A (1, 1)
2 B M (2, 4)
Q.36
2x + y = k
M lies on the line 2x + y = k 8 14 k=2· = 6. 5 5
Q.35
Statement 1 : An equation of a common tangent to the parabola y2 = 16 3 x and the
Ans. Sol.
Three numbers are chosen at random without replacement from {1, 2, 3, ........, 8}. The probability that their minimum is 3, given that their maximum is 6, is (1)
1 4
(2)
(3)
3 8
(4*)
4 3 , m (m 0) is a common tangent to the parabola
4
P(E) =
Ans.
y2 = 16 3 x and the ellipse 2x2 + y2 = 4, then m satisfies m4 + 2m2 = 24. (1) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 2. (2) Statement 1 is true, Statement 2 is false. (3) Statement 1 is false, Statement 2 is true. (4*) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. [4]
Sol.
Equation of tangent to parabola y2 = 16 3 x is y = mx + to ellipse
Q.37
4 3 . A line y = mx + C is tangent m
x 2 y2 = 1. When c2 = a2m2 + b2. a 2 b2
6 3 = 2m2 + 4 m2 m4 + 2m2 = 24 On solving, m2 = 4 or m2 = – 6 (not possible)] m=±2
T : y = 2x + 2 3 & y = – 2x – 2 3 .
Ans. Sol.
1 5
[4] S : {1, 2, 3, 4, 5, 6, 7, 8}
ellipse 2x2 + y2 = 4 is y = 2x + 2 3 . Statement 2 : If the line y = mx +
2 5
6
1 C3 = . 5 C3
Let ABCD be a parallelogram such that AB q , AD p and BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by p·q (1) r q p p·p 3 p · q (2) r 3q p p·p 3 p · q (3) r 3q p p·p p·q (4*) r q p p·p [4] r p q D r · p p q · p 0 = p·pp·q p A p·q = p·p p·q r q p . p·p
C r q
B
Q.38
Ans. Sol.
An equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is (1) x – 2y + 2z – 1 = 0 (2) x – 2y + 2z + 5 = 0 (3*) x – 2y + 2z – 3 = 0 (4) x – 2y + 2z + 1 = 0 [3] Equation of plane parallel to x – 2y + 2z – 5 = 0 is x – 2y + 2z = . Distance from origin is 1. |000|
Q.39
Ans. Sol.
Q.40
If f : R R is a function defined by
Ans. Sol.
2x 1 f(x) = [x] cos , where [x] denotes 2 the greatest integer function, then f is (1) discontinuous only at non-zero integral values of x. (2) continuous only at x = 0. (3*) continuous for every real x. (4) discontinuous only at x = 0. [3] f:RR
=1 12 22 22 = ± 3 P : x – 2y + 2z = ± 3.
2x 1 f(x) = [x] cos , 2 [ ] greatest integer function When x I, then f(x) = 0
In a PQR, if 3 sin P + 4 cos Q = 6 and 4sin Q + 3cos P = 1, then the angle R is equal to
2x 1 [ cos = 0 for n I] 2 For x I then f(x) is product of two continuous function therefore it is continuous. f(x) is continuous for every real x.
3 4
(1)
4
(2)
(3)
5 6
(4*)
6
Q.41
[4] 3sin P + 4cos Q = 6 ......(1) 4sin Q + 3cos P = 1 ......(2) Square and add (1) & (2) 24 sin (P + Q) = 12 1 sin (P + Q) = 2
P+Q=
But when P + Q =
n
Statement 2 :
R
then (1) & (2) not 6
satisfied 5 P+Q= R= . 6 6
k
3
(k 1)3 = n3, for any
k 1
P
5 Q or 6 6
Statement 1 : The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 +12 + 16) + ...... + (361 + 380 + 400) is 8000.
Ans.
natural number n. (1) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 2. (2) Statement 1 is true, Statement 2 is false. (3) Statement 1 is false, Statement 2 is true. (4*) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. [4]
Sol.
S-1: 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 +12 + 16) + ...... + (361 + 380 + 400) Clearly number of terms in sequence are 20. n
S-2 :
k
3
Sol.
(k 1)3 = n3 is true.
k 1
1 0 0 x 1 2 1 0 y = 0 3 2 1 z 0
The sum of given series for 20 terms is (20)3 = 8000. Q.42
The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is (1)
6 5
(3*) Ans. Sol.
`
10 3
(2)
5 3
(4)
3 5
Radius =
Q.43
1 u1 = 2 ; u2 = 1
Q.44
5 10 . Diameter = . 3 3
1 0 0 Let A = 2 1 0 . If u1 and u2 are column 3 2 1 1 0 matrices such that Au1 = 0 and Au2 = 1 , 0 0
Ans. Sol.
Ans.
1 1 (2*) 1
1 1 (3) 0
1 1 (4) 1
[2]
If n is a positive integer, then 2n
2n
3 1 3 1 is (1) an even positive integer (2) a rational number other than positive integers (3*) an irrational number (4) an odd positive integer [3] If n = 1, 2
2
3 1 3 1 = 4 3 . An irrational number.
Q.45
then u1 + u2 is equal to 1 1 (1) 0
0 1 2
1 u1 + u2 = 1 . 1
[3] (x – h)2 + (y – k)2 = k2 Centre (h, k), Radius = k 5 (1 – h)2 + k2 = k2 } h = 1 ; k = 3
x u1 = y z
Ans. Sol.
Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is (1) 630 (2*) 879 (3) 880 (4) 629 [2] (11) (10) (8) – 1 = 879.
Q.46
Ans. Sol.
An ellipse is drawn by taking a diameter of the circle (x – 1)2 + y2 = 1 as its semi-minor axis and a diameter of the circle x2 + (y – 2)2 = 4 as its semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is (1) 4x2 + y2 = 8 (2*) x2 + 4y2 = 16 2 2 (3) 4x + y = 4 (4) x2 + 4y2 = 8 [2] a = 4, b = 2
Q.48
x y2 =1 16 4 x2 + 4y2 = 16.
If the line
x 1 y 1 z 1 2 3 4
and
Ans. Sol.
x 3 yk z intersect, then k is equal to 1 2 1 9 2
(2) 0
(3) – 1
2 (4) 9
(1*)
Ans.
[1]
Sol.
If the line
x 1 y 1 z 1 2 3 4
x 3 yk z 1 2 1
intersect,
1 1 and b = . 2 4 (1) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 2. (2) Statement 1 is true, Statement 2 is false. (3) Statement 1 is false, Statement 2 is true. (4*) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. [4] f(x) = ln | x | + bx2 + ax, x 0
Statement 2 : a =
Ellipse
Q.47
Let a, b R be such that the function f given by f(x) = ln | x | + bx2 + ax, x 0 has extreme values at x = – 1 and x = 2. Statement 1 : f has local maximum at x = – 1 and at x = 2.
1 + 2bx + a x extreme values at x = – 1, 2 – 1 – 2b + a = 0 a – 2b = 1
f '(x) =
and
1 1 + 4b + a = 0 a + 4b = ...(2) 2 2
From (1) and (2) a = and
...(1)
1 1 ,b= 2 4
1 1 1 = 2 x x f "(–1) < 0 and f "(2) < 0 f has local maximum at x = – 1 and x = 2.
again f "(x) = 2b – then
3 1 k 1 0 1 2 1
3 2
4 1
=0
Q.49
2 k 1 1 3 4 =0 2 1 2 1
2 (3 – 8) – (k + 1) (2 – 4) – 1 (4 – 3) = 0 – 10 + 2k + 2 – 1 = 0 2k = 9 k =
9 . 2
Ans. Sol.
z is real, then the point z 1 represented by the complex number z lies (1) either on the real axis or on a circle not passing through the origin. (2) on the imaginary axis. (3*) either on the real axis or on a circle passing through the origin. (4) on a circle with centre at the origin. [3] If z 1and
z2 is real z 1 Let z = x + iy
x 2 y 2 2ixy z2 = ( x 1) iy z 1 =
0
( x 2 y 2 2ixy) ( x 1 iy) ( x 1) 2 y 2
Ans. Sol.
The negation of the statement "If become a teacher, then I will open a school", is (1) Neither I will become a teacher nor I will open a school. (2) I will not become a teacher or I will open a school. (3*) I will become a teacher and I will not open a school. (4) Either I will not become a teacher or I will not open a school. [3] P = I be one a teacher. Q = I will open a school. ~ (p q) = P ^ ~ q I become a teacher and I will not open a school. x
Q.51
If g(x) = cos 4t dt , then g(x + ) equals 0
(1*) g(x) – g()
Ans.
(2) g(x) · g()
g( x ) (3) g () [1], [4]
(4*) g(x) + g()
Given g(x) = cos 4t dt 0
Q.52
cos 4t dt 0
0
= g() + g(x) g(x + ) = g(x) + g() but g() = 0 g (x + ) = g (x) + g () = g (x) – g() A spherical balloon is filled with 4500 cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is 2 9
(1*) (3)
9 7
Ans.
[1]
Sol.
Given
(2)
9 2
(4)
7 9
dV = 72 dt
t=0 Volume of gas = 4500 t = 49 minute, Volume of gas = 72 × 49 = 3528 After 49 minute volume of gas inside balloon = (4500 – 3528) = 972
4 3 r = 972 3
Now, V =
4 3 r 3
– 72 = 4(92)
x
Now, g(x + ) =
x
0
dV dr = 4r2 dt dt
x
Sol.
Now, imaginary part of
Q.50
cos 4 t dt
= cos 4 t dt cos 4 t dt
2
z equal to zero. z 1 – y (x2 – y2) + (x – 1) 2xy = 0 y (x2 + y2 – 2x) = 0 y = 0 or x2 + y2 – 2x = 0.
x
= cos 4 t dt
dr 2 = . dt 9
dr dt
r = 9m
Q.53
Ans. Sol.
The equation esinx – e–sin x – 4 = 0 has (1) exactly one real root. (2) exactly four real root. (3) infinite number of real roots. (4*) no real roots. [4] Let esin x = k k–
Ans. Sol.
2 20 2 y dy = Area = 2 9 y . 3 4 0
Ans. Sol.
Q.55
Let X = {1, 2, 3, 4 , 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y X, Z X, and Y Z is empty, is (1) 25 (2) 53 2 (3) 5 (4*) 35 [4] 1 can be distributed in two set Y and Z by 3 ways. 2 can be distributed in two set Y and Zby 3 ways. 3 can be distributed in two set Y and Z by 3 ways. 4 can be distributed in two set Y and Z by 3 ways. 5 can be distributed in two set Y and Z by 3 ways. (The three ways are either only in Y or only in Z or in none of Y and Z.) Number of way of distributing each element = 35.
B
Q.56
Ans. Sol.
Q.57
y and x2 = 9y, and the straight line y = 2 4
is (1*)
20 2 3
(3) 20 2
(2) 10 2 (4)
10 2 3
x2 9
y=2 x
The area bounded between the parabolas x2 =
y=
A
k = esin x = 2 + 5 = It is greater than e Not possible
Q.54
y=4x2
y
1 – 4 = 0 k2 – 4k – 1 = 0 k
and esin x = 2 – 5 = negative i.e. not possible. No solution.
[1]
Ans.
Let P and Q be 3 × 3 matrices with P Q. If P3 = Q3 and P2Q = Q2P, then determinant of (P2 + Q2) is equal to (1*) 0 (2) – 1 (3) – 2 (4) 1 [1] P3 = Q3 P2Q = Q2P P3 – P2Q = Q3 – Q2P P2 (P – Q) = Q2 (Q – P) (P2 + Q2) (P – Q) = 0 | P 2 + Q2 | | P – Q | = 0 | P2 + Q2 | = 0 or | P – Q | = 0. Let x1, x2, ........., xn be n observations, and let x be their arithmetic mean and 2 be their variance. Statement 1: Variance of 2x1, 2x2, ...., 2xn is 42. Statement 2: Arithmetic mean of 2x1, 2x2, ...., 2xn is 4 x . (1) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1. (2*) Statement 1 is true, Statement 2 is false. (3) Statement 1 is false, Statement 2 is true. (4) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. [2]
Sol.
S-2 :
Q.59
Arithmetic Mean =
2 x1 2 x 2 ........ 2 x n n
x1 x 2 ....... x n = 2x = 2 n Statement-2 is false Statement-1 We know variance of x1, x2, x3, ....... xn.
x =
2 i
x i n n variance of 2x1 ; 2x2 ; ............ ; 2xn 2 =
4 x i2 n
2
A line is drawn through the point (1, 2) to meet the coordinates axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ, is (1*) – 2 (3)
(2)
1 4
Ans.
[1]
Sol.
Area =
1 2
(4) – 4
1 2 1 ( 2 m ) 2 m y
x 4 i n
=
1 4 m 4 2 m
dA =0 dx
x i2 x i = 42. = 4 n n
(0, 2 – m) (1, 2) O
Gives m 2 = 4 m = ± 2
2 1 , 0 m
x
Area m = – 2. Q.58
The population p(t) at time t of a certain mouse species satisfies the differential equation
Q.60
dp( t ) = 0.5 p(t) – 450. If p(0) = 850, then dt the time at which the population becomes zero is 1 ln 18 2 (3*) 2 ln 18 [3]
(1) Ans. Sol.
(2) ln 18 (4) ln 9
dp( t ) 1 = p(t) – 450 dt 2
If e
1 dt 2
= e–t/2
p(t) e–t/2 = – 450 e t 2dt + k p(t) = 900 + ket/2 .......(1) When t = 0 ; P(0) = 850 k = – 50 Equation (1) becomes p(t) = 900 – 50et/2 when p(t) = 0 then t = 2ln 18.
Ans. Sol.
If 100 times the 100th term of an AP with non zero common difference equals the 50 times its 50th term, then the 150th term of this AP is (1) 150 (2*) zero (3) – 150 (4) 150 times its 50th term [2] 100 (a + 99d) = 50 (a + 49d) a + 149d = 0 .......(1) and T150 = a + 149d = 0 [From (1)].
Q.61
Ans. Sol.
This question has Statement-1 and statement2. Of the four choices given after the Statements, choose the one that best describes the two statements. If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2. Statement-1 : If stretched by the same amount, work done on S1, will be more than that on S2. Statement -2 :k1 < k2 (1) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is false and Statement-2 is not the correct explanation of Statement-1 (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false [1] Stretched by same force hence k1x1 = k2x2 More work is done on spring-1 hence
Q.62
This question has Statement-1 and statement2. Of the four choices given after the Statements, choose the one that best describes the two statements. An insulating solid sphere of radius R has uniformly positive charge density . As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero. Statement-1 : When a charge 'q' is taken from the centre to the surface of the sphere, q its potential energy changes by 3 . 0
Statement -2 :The electric field at a distance r (r < R) from the centre of the sphere is r 3 0 .
Ans.
(1) Statement-1 is false, Statement-2 is true. (2) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of Statement-1. (3) Statement-1 is true, Statement-2 is false and Statement-2 is not the correct explanation of Statement-1 (4) Statement-1 is true, Statement-2 is false [1]
Sol.
Vanter =
1 1 k1x12 > k2x22 2 2 x1 > x2 k1 < k2
3kQ kQ , Vsurface = 2R R
R 2 q qkQ U = qV = = 6 0 2R
Q.63
A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area S and length L. L is lightly less than 2R. To fit the ring on the wheel, it is heated so that its temperature rises by T and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is , and its Youngs' modulus is Y, the force that one part of the wheel applies on the other part is :
Q.65
Ans. Sol. Q.66
Ans.
A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance up to which it can detect object located on the surface of the earth (Radius of earth = 6.4 × 106 m) (1) 40 km (2) 64 km (3) 80 km (4) 16 km [3]
Sol.
d=
Q.67
Truth table for system of four NAND gates as shown in figure is :
R
Ans. Sol.
(1) SYT (2) 2SYT (3) 2SYT (4) SYT [2] Thermal stress = YT Force developed = YST
Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be (1) 5 (2) 6 (3) 2 (4) 3 [2] No. of spectral lines = nC2 = 4C2 = 6
2 6.4 103 0.5 = 80 km
2Rh =
A
T
force exerted by the part = 2YST
Y
T
Q.64
B
A diatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by : 2n 2 h 2
(1)
(m1 m 2 )r
(2)
2
2 2 2
(3) Ans. Sol.
(m1 m 2 ) n h 2m12 m 22 r 2
(m1 m 2 )n 2 h 2 2m1m 2 r
2
Putting we get E =
(3)
2 2
n h
(4)
2(m1 m 2 )r 2
[2] I n Rotational energy =
(1)
1 2 I where I = µr2 2
n 2 (m1 m 2 ) 2m1m 2 r 2
Ans.
[3]
A 0 0 1 1
B 0 1 0 1
Y 1 1 0 0
A 0 0 1 1
B 0 1 0 1
Y 0 1 1 0
(2)
(4)
A 0 0 1 1
B 0 1 0 1
Y 1 0 0 1
A 0 0 1 1
B 0 1 0 1
Y 0 0 1 1
Q.68
Ans.
A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : 58.5 degree Vernier scale reading: 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data: (1) 58.65 degree (2) 59 degree (3) 58.59 degree (4) 58.77 degree [1]
Sol.
Least count (LC) =
Q.70
E
0 .5 30 = 58.65 degree
(2) R
Ans.
This question has Statement-1 and statement-2. Of the four choices given after the Statements, choose the one that best describes the two statements. Statement-1 : Davisson - Germer experiment established the wave nature of electrons. Statement-2 : If electrons have wave nature, they can interfere and show diffraction. (1) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is false and Statement-2 is not the correct explanation of Statement-1 (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false [1]
r
E
r
R
r
(4) R
Ans.
[1]
Sol.
Einside r
Q.71
R E
(3)
Eoutsides Q.69
E
(1)
0.5 deg ree 30 Reading = Main scale reading + vernier scale reading
= 58.5 + 9 ×
In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be :
r
1 r2
Ans.
A cylindrical tube, open at both ends, has a fundamental frequency, f, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now: 3f (1) (2) 2f 4 f (3) f (4) 2 [3]
Sol.
f=
v ; = 2L
v L ; = = 2L 4 2 hence f ' = f
f'=
Ans. Sol.
If a simple pendulum has Significant amplitude (up to a factor of 1/ e of original) only in the period between t = 0s to t = s, then may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with ‘b’ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds: (1)
1 b
(2)
(3)
0.693 b
(4) b
Q.74
[1] a = –bv hence v = v 0 e–bt
1 1 Average life time = = b
Q.73
Ans.
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to : (1) shielding of magnetic lines of force as aluminium is a paramagnetic material. (2) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping. (3) development of air current when the plate is placed. (4) induction of electrical charge on the plate [2]
25 20 15 10 5 0
2 b
comparing with N = N0e–t
Potential difference V in volts
Q.72
Ans. Sol.
50 100 150 200 250 300 Time t in seconds
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant of this circuit lies between: (1) 50 sec and 100 sec (2) 100 sec and 150 sec (3) 150 sec and 200 sec (4) 0 and 50 sec [2] V = Ve–t/
at t = 200 sec, V = 5, V0 = 25 hence we get = 124.2 sec Q.75
Ans.
A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be : (1) 750 K (2) 600 K (3) efficiency of Carnot engine cannot be made larger than 50% (4) 1200 K [1]
Sol.
0.4 = 1 –
Tsin k Tsink = 300 K 500
300 0.6 = 1 – T Tsource = 750 K source
Q.76
Ans.
Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440V supply. Which of the bulbs will fuse? (1) 25W (2) neither (3) both (4) 100 W [1] R
Q.79
In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference is given by :
4R
100W
25W
(1)
Im 2 1 4 cos 5 2
(2)
Im 2 1 8 cos 9 2
(3)
Im 4 5 cos 9
(4)
Im 2 1 2 cos 3 2
Sol. 440V
Q.77
Ans. Q.78
Ans. Sol.
An electromagnetic wave in vacuum has the electric and magnetic fields E and B , which are always perpendicular to each other. The direction of polarization is given by X and that of wave propagation by k . Then (1) X || B and k || E B (2) X || E and k || B E (3) X || B and k || B E (4) X || E and k || E B [4] The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g' and 'R' (radius of earth) are 10 m/s 2 and 6400 km respectively. The required energy for this work will be : (1) 6.4 × 109 Joules (2) 6.4 × 1010 Joules (3) 6.4 × 1011 Joules (4) 6.4 × 108 Joules [2] GmMe On surface of earth U = – mgRe Re = –6.4 × 1010 Joule
Ans. Sol.
[2] I0 and 4I0 Im = 9I0 Ires = I0 + 4I0 + 2 4I 02 cos = 5I0 + 4I0 cos =
Q.80
Ans. Sol.
Im 2 1 8 cos 2 9
A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be : (1) 10 2 m
(2) 20 m
(3) 20 2 m [2]
(4) 10 m
u2 = 10 m 2g u2 Rmax = = 20 m g
Q.82
Ans. Sol.
t
An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film ? (1) 3.2 m (2) 5.6 m (3) 7.2 m (4) 2.4 m [2]
(3)
0
* New v should be = 12 –
0
[3]
Sol.
dT = –k(T – Ts) dt
ln
*
A liquid in a beaker has temperature (t) at time t and 0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge( – 0) and t is:
(1)
T Ts kt T0 Ts
ln(T – Ts) = ln(T0 – Ts) – kt Q.84
Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas) 2P0
loge ( – 0)
Q.83
t
Ans.
1 35 = CM 3 3
1 1 1 3 7 1 = = =– u v f 35 80 500 u = –5.6 m
t
(4)
1 1 1 1 7 + = = 12 240 f f 80
1 1 x due to slab = t 1 = CM 3
loge ( – 0)
(2)
loge ( – 0)
Ans.
Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (Mass of neutron = 1.6725 × 10–27 kg Mass of proton = 1.6725 × 10–27 kg Mass of electron = 9 × 10–31 kg) (1) 6.30 MeV (2) 5.4 MeV (3) 0.73 MeV (4) 7.10 MeV NO ANSWER (WRONG DATA) but correct answer is [3] with actual data.
loge ( – 0)
Q.81
P0
B
A V0
t
(1) 10.5% (3) 15.4%
C
D
2V0
(2) 12.5% (4) 9.1%
Q.87
Ans.
[3]
Sol.
p0 v0 2P0 v 0 QAB = nCv , QBC = nCp R R
2P0 v 0 p0 v0 QCD = nCv , QDA = nCp R R Qrepected = Q0 + QDA Qabsorbed = QAB + QBC
Q rej
=1– Q.85
Ans.
Qab
A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F (t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves? F0 mb
(1) v(t) t
= 0.154 F0 mb
Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively rp, rd and r. Which one of the following relations is correct? (1) r > rd > rp (2) r = rd > rp (3) r = rp = rd (4) r = rp < rd [4]
(2) v(t) t F0 mb
(3) v(t) t
Sol.
mv R= = qB
2mk qB
RP : Rd : R = =1:
Q.86
m : e
F0b m
2m : e
4m 2e
(4) v(t) t
2 :1
Ans.
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is : (1) 1% (2) 3% (3) 6% (4) zero [3]
Sol.
R=
V I
R V I = + = 6% R V I
Ans.
[1]
Sol.
f 0 e bt a= m v
t
dv =
m0 e
0
v=
f
bt
0
f0 (1 e bt ) mb
dt
Q.88
A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 × 10– 2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is:
(1)
B
R
FILM
(2)
B
R
w
Ans. Sol.
Q.89
Ans. Sol.
(1) 0.05 Nm–1 (3) 0.0125 Nm–1 [2] 2SL = weight S = 0.025 N/m
(2) 0.025 Nm–1 (4) 0.1 Nm–1
Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is : (1) r1 : r2 (2) 1 : 1 (3) m1r1 : m2 r2 (4) m1 : m2 [1] a1 r1 r12 = = a2 r2 r12
(3)
B
R
(4)
B
R
Ans.
[3]
Sol.
dB =
µ 0 dI 2r
dr r
Integrating we get B Q.90
A charge Q is uniformly distributed over the surface of non-conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with a angular velocity . As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure. :
dq = (2r dr) = dI =
1 R
2Qrdr R2
Qrdr (dq) = 2 R 2
